On Tue, Jun 29, 2010 at 5:56 AM, Lawrence D'Oliveiro
<l...@geek-central.gen.new_zealand> wrote:
> In message <mailman.2231.1277700501.32709.python-l...@python.org>, Kushal
> Kumaran wrote:
>
>> On Sun, Jun 27, 2010 at 5:16 PM, Lawrence D'Oliveiro
>> <l...@geek-central.gen.new_zealand> wrote:
>>
>>>In message <mailman.2184.1277626565.32709.python-l...@python.org>, Kushal
>>> Kumaran wrote:
>>>
>>>> On Sun, Jun 27, 2010 at 9:47 AM, Lawrence D'Oliveiro
>>>> <l...@geek-central.gen.new_zealand> wrote:
>>>>
>>>>> A long while ago I came up with this macro:
>>>>>
>>>>> #define Descr(v) &v, sizeof v
>>>>>
>>>>> making the correct version of the above become
>>>>>
>>>>> snprintf(Descr(buf), foo);
>>>>
>>>> Not quite right. If buf is a char array, as suggested by the use of
>>>> sizeof, then you're not passing a char* to snprintf.
>>>
>>> What am I passing, then?
>>
>> Here's what gcc tells me (I declared buf as char buf[512]):
>> sprintf.c:8: warning: passing argument 1 of ‘snprintf’ from
>> incompatible pointer type
>> /usr/include/stdio.h:363: note: expected ‘char * __restrict__’ but
>> argument is of type ‘char (*)[512]’
>>
>> You just need to lose the & from the macro.
>
> Why does this work, then:
>
> l...@theon:hack> cat test.c
> #include <stdio.h>
>
> int main(int argc, char ** argv)
> {
> char buf[512];
> const int a = 2, b = 3;
> snprintf(&buf, sizeof buf, "%d + %d = %d\n", a, b, a + b);
> fprintf(stdout, buf);
> return
> 0;
> } /*main*/
> l...@theon:hack> ./test
> 2 + 3 = 5
>
By accident. I hope your compiler warned you about your snprintf call.
Reading these threads might help you understand how char* and char
(*)[512] are different:
http://groups.google.com/group/comp.lang.c++/browse_thread/thread/24708a9204061ce/848ceaf5ec774d81
http://groups.google.com/group/comp.lang.c.moderated/browse_thread/thread/fe264c550947a2e5/32b330cdf8aba3d6
--
regards,
kushal
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